Example 1 : <\/span>Find the range of the given function \\(log_{\\sqrt{2}}(2-log_2(16sin^2x+1))\\)<\/p>\n Solution : <\/span>Now 1 \\(\\le\\) \\(16sin^2x\\) + 1) \\(\\le\\) 17 Example 2 : <\/span> Find the inverse of the function f(x) = \\(log_a(x + \\sqrt{(x^2+1)})\\); a > 1 and assuming it to be an onto function.<\/p>\n Solution : <\/span>Given f(x) = \\(log_a(x + \\sqrt{(x^2+1)})\\) Example 3 : <\/span>Find the period of the function f(x) = \\(e^{x-[x]+|cos\\pi x|+|cos2\\pi x|+ ….. + |cosn\\pi x|}\\)<\/p>\n Solution : <\/span>f(x) = \\(e^{x-[x]+|cos\\pi x|+|cos2\\pi x|+ ….. + |cosn\\pi x|}\\) Practice these given function examples to test your knowledge on concepts of function.<\/p>\n \n <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":" Here you will learn some function examples for better understanding of function concepts. Example 1 : Find the range of the given function \\(log_{\\sqrt{2}}(2-log_2(16sin^2x+1))\\) Solution : Now 1 \\(\\le\\) \\(16sin^2x\\) + 1) \\(\\le\\) 17 \\(\\therefore\\) 0 \\(\\le\\) \\(log_2(16sin^2x+1)\\) \\(\\le\\) \\(log_217\\) \\(\\therefore\\) 2 – \\(log_217\\) \\(\\le\\) 2 – \\(log_2(16sin^2x+1)\\) \\(\\le\\) 2 Now consider 0 …<\/p>\n
\n \\(\\therefore\\) 0 \\(\\le\\) \\(log_2(16sin^2x+1)\\) \\(\\le\\) \\(log_217\\)
\n \\(\\therefore\\) 2 – \\(log_217\\) \\(\\le\\) 2 – \\(log_2(16sin^2x+1)\\) \\(\\le\\) 2
\n Now consider 0 < 2 – \\(log_2(16sin^2x+1)\\) \\(\\le\\) 2
\n \\(\\therefore\\) -\\(\\infty\\) < \\(log_{\\sqrt{2}}(2-log_2(16sin^2x+1))\\) \\(\\le\\) \\(log_{\\sqrt{2}}2\\) = 2
\n \\(\\therefore\\) the range is (-\\(\\infty\\), 2]<\/p>\n
\n\n \n
\n \\(\\therefore\\) f'(x) = \\(log_ae\\over {\\sqrt{1+x^2}}\\) > 0
\n which is strictly increasing functions.
\n Thus, f(x) is injective, given that f(x) is onto. Hence the given function f(x) is invertible.
\n Interchanging x & y
\n \\(\\implies\\) \\(log_a(y + \\sqrt{(y^2+1)})\\) = x
\n \\(\\implies\\) \\(y + \\sqrt{(y^2+1)}\\) = \\(a^x\\) ……..(1)
\n and \\(\\sqrt{(y^2+1)}\\) – y = \\(a^{-x}\\) ………..(2)
\n From (1) and (2), we get y = \\(1\\over 2\\)(\\(a^x – a^{-x}\\)) or \\(f{-1}\\)(x) = \\(1\\over 2\\)(\\(a^x – a^{-x}\\)).\n <\/p>
\n\n \n
\n Period of x – [x] = 1
\n Period of \\(|cos\\pi x|\\) = 1
\n Period of \\(|cos2\\pi x|\\) = \\(1\\over 2\\)
\n ……………………………….
\n Period of \\(|cosn\\pi x|\\) = \\(1\\over n\\)
\n So period of f(x) will be L.C.M of all period = 1.<\/p>\n